4569846498
Modelling and analysis of suspension systems 187 Force (N) 7000.0 6500.0 6000.0 5500.0 5000.0 4500.0 POTHOLE BRAKING LOADCASE Magnitude of force in the spring (static at time 0 to full load at time 1.0) 4000.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Time (s) Fig. 4.50 MSC.ADAMS plot of spring load for pothole braking case 4.8.3 Case study 3 – Dynamic durability loadcase In this case study we extend the model of the single suspension system to include an additional part representing the corner of the vehicle body, quarter model, to which the suspension linkages attach as shown in Figure 4.51. The vehicle body is attached to the ground part by a translational joint that allows the body to move vertically in response to the loads transmitted through the suspension system. A jack part is reintroduced to reproduce vertical motion inputs representative of road conditions. An additional complexity in the model is the introduction of stiffness and damping terms Quarter vehicle body part Body connects to ground by translational joint Tyre representation by spring and damper forces Jack part Motion input to represent road profile Fig. 4.51 Quarter vehicle body and suspension model
188 Multibody Systems Approach to Vehicle Dynamics in a force element that represents the behaviour of the tyre. The force element acts between the centre of the wheel and a point on the jack coincident with the centre of the tyre contact patch. In this case this is assumed to be directly below the wheel centre. It is important that the force element for the tyre acts only in compression and allows the tyre to lift off the top of the jack if the input is severe enough. The tyre is taken to have linear radial stiffness of 160 N/mm and a damping coefficient of 0.5 Ns/mm. The undeformed radius of the tyre, equivalent to the free length of the tyre spring, is taken here to be 318 mm. A step function can also be used to zero the tyre force if contact with the jack is lost. The full definition can be accomplished using the following SFORCE statement that is taken to act between Marker 1010 at the wheel centre and Marker 2029 on the jack part: SFORCE/1029, I1010, J2029, TRANS ,FUSTEP(318DM(1010,2029),0,0,0.1,160*(318DM(1010,2029)) 0.5*VR(1010,2029)) The next step is to define the motion imparted to the jack part to represent the input from the road surface. This is illustrated in Figure 4.52 where the profile for a ‘sleeping policeman’ road obstacle is given. The profile is defined as a set of xy pairs. Note here that the xy values are local to the definition of the obstacle profile and not associated with the X- and Y-axes of the ground reference frame. The vehicle is assumed to be moving with a forward speed of 10 m/s so that the x values associated with distance can be converted to time. 10 m/s Z X GRF 1000 200 200 200 200 200 200 10 000 y x 1 2 3 4 5 6 100 7 8 9 Point 1 2 3 4 5 6 7 8 9 Distance x (mm) Time x (s) Height y (mm) 0 1000 1200 1400 1600 1800 2000 2200 12 200 0 0.10 0.12 0.14 0.16 0.18 0.20 0.22 1.22 0 0 50 100 100 100 50 0 0 Fig. 4.52 Road profile for ‘sleeping policeman’ speed bump
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188 Multibody Systems Approach to Vehicle Dynamics<br />
in a force element that represents the behaviour of the tyre. The force element<br />
acts between the centre of the wheel and a point on the jack coincident<br />
with the centre of the tyre contact patch. In this case this is assumed to be<br />
directly below the wheel centre. It is important that the force element for<br />
the tyre acts only in compression and allows the tyre to lift off the top of the<br />
jack if the input is severe enough.<br />
The tyre is taken to have linear radial stiffness of 160 N/mm and a damping<br />
coefficient of 0.5 Ns/mm. The undeformed radius of the tyre, equivalent to<br />
the free length of the tyre spring, is taken here to be 318 mm. A step function<br />
can also be used to zero the tyre force if contact with the jack is lost.<br />
The full definition can be accomplished using the following SFORCE<br />
statement that is taken to act between Marker 1010 at the wheel centre and<br />
Marker 2029 on the jack part:<br />
SFORCE/1029, I1010, J2029, TRANS<br />
,FUSTEP(318DM(1010,2029),0,0,0.1,160*(318DM(1010,2029))<br />
0.5*VR(1010,2029))<br />
The next step is to define the motion imparted to the jack part to represent the<br />
input from the road surface. This is illustrated in Figure 4.52 where the profile<br />
for a ‘sleeping policeman’ road obstacle is given. The profile is defined as<br />
a set of xy pairs. Note here that the xy values are local to the definition of the<br />
obstacle profile and not associated with the X- and Y-axes of the ground reference<br />
frame. The vehicle is assumed to be moving with a forward speed of<br />
10 m/s so that the x values associated with distance can be converted to time.<br />
10 m/s<br />
Z<br />
X<br />
GRF<br />
1000 200 200 200 200 200 200 10 000<br />
y<br />
x<br />
1<br />
2<br />
3<br />
4<br />
5 6<br />
100<br />
7<br />
8 9<br />
Point<br />
1<br />
2 3 4 5 6 7 8 9<br />
Distance x (mm)<br />
Time x (s)<br />
Height y (mm)<br />
0 1000 1200 1400 1600 1800 2000 2200 12 200<br />
0 0.10 0.12 0.14 0.16 0.18 0.20 0.22 1.22<br />
0 0 50 100 100 100 50 0 0<br />
Fig. 4.52<br />
Road profile for ‘sleeping policeman’ speed bump